# Chapter 6. Dynamics I: Motion Along a Line

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Chapter 6. Dynamics I: Motion Along a Line
Chapter Goal: To learn how to solve problems about motion in a straight line.

Student Learning Objectives – Ch. 6
• To draw and make effective use of free-body diagrams. • To recognize and solve simple equilibrium problems. • To distinguish mass, weight, and apparent weight. • To learn and use simple models of friction. • To apply the full strategy for force and motion problems to problems in single-particle dynamics.

Equilibrium An object on which the net force is zero is said to be in equilibrium. Static equilibrium: object is at rest. Dynamic equilibrium: moving along a straight line with constant velocity. Both are identical from a Newtonian perspective because the net force and the acceleration are zero.

Problem-Solving Strategy: Equilibrium Problems

Problem-Solving Strategy: Equilibrium Problems

Example Equilibrium Problem (#31)
A 500 kg piano is being lowered into position by a crane while 2 people steady it with ropes pulling to the side. Bob’s rope pulls left, 150 below horizontal, with 500 N of tension. Ellen’s rope pulls right, 250 below horizontal. What tension must Ellen maintain in her rope to keep the piano descending at a steady speed? What is the tension in the main cable?

Example Problem (#31) – Freebody Diagram
T3 T1 is Bob’s side, T2 is Ellen’s side, T3 is the main cable. Same system for the angles

Problem #31 - Apply Newton’s 1st Law
ΣFy = 0, ΣFx = 0

Problem #31 – Solve and assess
T2 = 533 N, T3 = 5.25 x 103 N. The cable must supports the weight of the piano (4900 N) plus the added downward components of the tension in the supporting ropes.

When your car rapidly accelerates forward, you are pressed into the seat. When your car suddenly comes to a halt, you lunge forward. These two phenomena are an example of: Newton’s 1st Law; the sum of the forces upon you is zero. Newton’s 2nd Law; you experience a net force and you accelerate Newton’s 3rd Law; you experience a force equal and opposite to that of the car

When your car rapidly accelerates forward, you are pressed into the seat. When your car suddenly comes to a halt, you lunge forward. These two phenomena are an example of: Newton’s 1st Law; the sum of the forces upon you is zero. Newton’s 2nd Law; you experience a net force and you accelerate Newton’s 3rd Law; you experience a force equal and opposite to that of the car

Problem-Solving Strategy for Newton’s 2nd Law Problems
Use the problem-solving strategy outlined for Newton’s 1st Law problems to draw the free body diagram and determine known quantities. Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. Reverse # 2 and 3 if necessary. 12

For which of the following is may = F1 -F2 cosθ - F3 sin θ
A B C D

For which of the following is may = F1 -F2 cosθ - F3 sin θ
A B C D

Example Dynamics Problem
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom?

Example Dynamics Problem – Visualize Freebody Diagram
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? n FG Find v1 a

Example Dynamics Problem – Newton’s 2nd Law in component form to solve for acceleration
A 75-kg skier starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? n ΣFy = may = 0 ΣFx = max a = 1.7 m/s2 Supports earlier statement that a = g sinθ FG

Example Dynamics Problem – Use kinematics to find speed
Example Dynamics Problem – Use kinematics to find speed. Is time important? A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? Note (the slope is 50 m high, not long!) v1 = 31.3 m/s. That’s about 60 mph! Find v1 a = 1.7 m/s2, from previous

Use the graph to answer the question #1
The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the acceleration of the object at t = 2s? 4 m/s2 2 m/s2 8 m/s2 0 m/s2

Use the graph to answer the question #1
The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the acceleration of the object at t = 2s? 4 m/s2 2 m/s2 8 m/s2 0 m/s2

Use the graph to answer the question #2
The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the velocity of the object at t = 6s? A. v = 4m/s B. v = -1 m/s v = 0 m/s v = 2 m/s

Use the graph to answer the question #2
The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the velocity of the object at t = 6s? A. v = 4m/s B. v = -1 m/s v = 0 m/s v = 2 m/s

Mass and Gravity Mass is a scalar quantity that describes the amount of matter in an object. Mass is an intrinsic property of an object. The force of gravity is an attractive, long-range “inverse square” force between any two objects.

The Earth and the Moon The figure shows the moon (m1) and the earth (m2). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F2on1 ). The moon also exerts a force on the earth, F1on2, shown in blue (not to scale!). The magnitude of this force is: a. about 80 smaller than F2on1 b. somewhat smaller than F2on1 c. Equal to F2on1 d. Not related to F2on1. moon ? STT12.2 Answer: C earth

The Earth and the Moon The figure shows the moon (m1) and the earth (m2). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F2on1 ). The moon also exerts a force on the earth, F1on2, shown in blue (not to scale!). The magnitude of this force is: Equal to F2on1 moon STT12.2 Answer: C earth

where the quantity g is defined to be
Consider an object of mass m, on or near the surface of a planet. We can write the gravitational force even more simply as: gravitational (weight )force where the quantity g is defined to be M, R represent the mass and radius of the planet. The weight force is not an intrinsic property of an object and does not have a unique value. The direction of the gravity vector defines true vertical.

Apparent Weight Apparent weight w, is a contact force (e.g. T, n, or Fsp), which can be thought of as “what the scale says”, although there is not always a scale. If object and scale are in vertical static or dynamic equilibrium w = FG = mg. If object and scale accelerate vertically, w ≠ mg. It must be calculated using Newton’s 2nd Law. The use of w instead of n or T is optional, as long as you know which force is the apparent weight.

An elevator suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is: less than your true weight, which is mg. equal to your true weight, which is mg. more than your true weight, which is mg. zero. Answer: C

An elevator suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is: less than your true weight, which is mg. equal to your true weight, which is mg. more than your true weight, which is mg. zero. n Answer: C

Workbook Problem # 18

Workbook Problem # 18 - ans
Explanation: S is the normal force which is the apparent weight. From N’s 2nd Law: S= ma + |mg| speed and direction are not relevant.

Apparent Weight Problem
A 50-kg woman gets in a 1000-kg elevator at rest. The elevator has a scale in it (I hate when that happens). As the elevator begins to move, the scale reads 600 N for the first 3 seconds. Can you tell which direction she moved? If so, what is it? How far has the elevator moved in those 3 s?

Apparent Weight Problem (a)
Calculate the true weight (mg) and draw a free-body diagram that correctly shows relative vector lengths In this case the scale reads heavy (I hate when that happens) so net force and acceleration are in the positive direction. Elevator must be going up and speeding up or going down and slowing down. Only one is possible due to initial velocity constraint. known: v0 = 0 m/s m = 50 kg n = 600 N g = 9.8 m/s2

Apparent Weight Problem (b)
How far has the elevator moved in those 3 s? The pictorial representation shows not enough information to solve problem, so we go to Newton’s 2nd Law analysis y1, t1 = 3s v1 a0 0 m y0 = t0 = v0 = 0

Apparent Weight Problem (b)
ΣFy = may using known values: n – mg = ma a = 2.2 m/s2 a known: m = 50 kg n = 600 N g = 9.8 m/s2

Apparent Weight Problem (b)
How far has the elevator moved in those 3 s? Time is important so use position equation with v0 = 0m/s: ∆ y = ½ a ∆t12 ∆ y = 9.9 m y1, t1 = 3s v1 a0 = 2.2 m/s2, determined from N’s 2nd Law 0 m y0 = t0 = v0 = 0

Friction

Kinetic Friction Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force. where the proportionality constant μk is called the coefficient of kinetic friction (table in text).

Static Friction The box is in static equilibrium, so the static friction must exactly balance the pushing force:

Static friction An object remains at rest as long as fs < fs max
The object slips when fs = fs max A static friction force fs > fs max is not physically   possible. fs max is always >fk for the same surfaces where the proportionality constant μs is called the  coefficient of static friction.

Comparison of static and kinetic friction

Rolling Friction Rolling friction acts much like kinetic friction, but values for ur are much less than those for uk. Rolling friction is a resistive force. It is not the same as the static friction that provides the propulsion force that move the wheel forward.

A model of friction “ motion” indicates motion relative to the two surfaces the max value static friction, fs max occurs at the very instant the object begins to move (which often means 1 ns before, for problem-solving purposes.

Rank in order, from largest to smallest, the magnitude of the friction forces in these five different situations. The box and the floor are made of the same materials in all situations. The push force is not necessarily the same. Answer: B

Rank order, from largest to smallest, the size of the friction forces in these five different situations. The box and the floor are made of the same materials in all situations. fb > fc = fd = fe > fa. STT5.3

Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope with μk = What is his speed at the bottom? This is the same problem as before only the slope is no longer frictionless. Before, the velocity was 31.3 m/s. How does friction change that?

Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope μs = 0.12 and μk = 0.06 What is his speed at the bottom? n fk FG Find v1

Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? v1 = 25.4 m/s. Compare with “frictionless” problem. Friction acts to slow him down, although not by much. Find v1 a, from previous

Example “max” problem A truck is hauling a crate when it starts up a 10.0˚ hill. The coefficients of friction are μs = 0.35, and μk = 0.15, respectively. What is the maximum acceleration the truck can have as he goes up the hill, without the crate slipping backward?

Example static friction problem
fs known find θ = 10˚ amax us = .35 uk = .15 θ Fg When does amax occur? Find n using Newton’s 2nd law in the y direction. Find amax using Newton’s 2nd law in the x direction.

Example static friction problem
fs known find θ = 10˚ amax us = .35 uk = .15 θ Fg amax =1.68 m/s/s

Keep the picture up A person is trying to judge whether a picture of mass 1.10 kg is properly positioned by pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between picture and wall is What is the minimum amount of pressing force required? Draw a freebody diagram. In which direction is the normal force in this problem? Does it have anything to do with the weight of the picture? 52

Keep the picture up Freebody diagram
fbd - picture Knowns m = 1.10 kg μs = 0.660 Find Fpush fs Fpush n Fg Forces which are usually x are y in this problem and vice versa (with the exception of gravity). Newton’s Laws still work. 53

Keep the picture up Freebody diagram
Knowns m = 1.10 kg μs = 0.660 Find Fpush fbd - picture fs Fpush Newton’s Law in the x direction tells us that n = Fpush but nothing else about the value of either. Moving right along to the y-direction: n Fg ΣFy = may = 0 = fs – FG or fs = mg. No matter how hard you press, the picture will not levitate up. Fact. However, if you don’t push hard enough…. 54

Keep the picture up - answer
Knowns m = 1.10 kg μs = 0.660 Find Fpush fbd - picture fs Fpush The minimum value of n must be the value that allows fsmax to be equal to the weight of the picture: ΣFy = 0 = fsmax – FG or μs |n| = mg n = Fpush = 16.3 N n FG 55

Using Newton’s 2nd Law: Workbook exercises Answers: 7-12